2015 AIME II Problems/Problem 7

Problem

Triangle $ABC$ has side lengths $AB = 12$ , $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial $$ (Error compiling LaTeX. Unknown error_msg)Area( $PQRS$ ) = \alpha \omega - \beta \omega^2.$$ (Error compiling LaTeX. Unknown error_msg)

Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

If $\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so

$\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0$

and $\alpha = 25\beta$ . If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over $PQ$ , $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$ ,

$[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90$

so

$45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}$

and

$\beta = \frac{180}{625} = \frac{36}{125}$

so the answer is $m + n = 36 + 125 = \boxed{161}$ .

Solution 2

$[asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]$

Similar triangles can also solve the problem.

First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$ )

After finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \frac{36}{5}$ . Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$ . We then know that $CE = \frac{77}{5}$ .

Now consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ , $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$ .

Let $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$ . Since $PF+FQ=PQ=\omega$ . We can solve for $PF$ and $FQ$ in terms of $\omega$ . We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$ .

Let's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$ . We can set up the proportion:

$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$ . Solving for $AF$ , $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$ .

We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$ .

Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$ .

This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$ .

Solution 3

Heron's Formula gives $[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,$ so the altitude from $A$ to $BC$ has length $\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.$

Now, draw a parallel to $AB$ from $Q$ , intersecting $BC$ at $T$ . Then $BT = w$ in parallelogram $QPBT$ , and so $CT = 25 - w$ . Clearly, $CQT$ and $CAB$ are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so $\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.$ Solving gives $[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}$ , so the answer is $36 + 125 = 161$ .

Solution 4

Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$ . Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$ , you find $\frac{w}{25}=\frac{h}{36/5}$ . Thus, $h=\frac{36w}{125}$ . To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$ , and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$ .

Solution 5

Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\frac{36}{5}$

Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$

Hence, $PS=\frac{3}{5}(12-L)$

Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$

Now, it is obvious that we want to find $L$ in terms of $W$ .

Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$

Hence.. we can now set up similar triangles.

We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$ .

Plugging back in..

$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$

Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$

Hence, $125+36=\boxed{161}$

Solution 6

Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$ , so $\beta = \frac{36}{125}$ , yielding $\boxed{161}$ .

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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